I am trying to find infimum of $\Big\{\frac{1}{\sqrt[n]{m}} + \frac{1}{\sqrt[m]{n}}, \quad n, m \in \mathbb{N}\Big\}$
First, by using geometric mean $\leq$ arithmetic mean, I get: $$\sqrt[n]{m} = \sqrt[n]{m \cdot 1^{n-1}} \leq \frac{m + (n - 1)}{n} \qquad n, m \in \mathbb{N}$$ $$\sqrt[m]{n} = \sqrt[m]{n \cdot 1^{m-1}} \leq \frac{n + (m - 1)}{m} \qquad n, m \in \mathbb{N}$$ now I invert: $$\frac{1}{\sqrt[n]{m}} \geq \frac{n}{m + (n - 1)} \qquad n, m \in \mathbb{N}$$ $$\frac{1}{\sqrt[m]{n}} \geq \frac{m}{n + (m - 1)} \qquad n, m \in \mathbb{N}$$ and sum both sides: $$\frac{1}{\sqrt[n]{m}}+\frac{1}{\sqrt[m]{n}}\geq \frac{n + m}{m + n - 1} \qquad n, m \in \mathbb{N}$$ conclude that $n + m > n + m - 1$ for $n, m \in \mathbb{N}$, therefore: $$\frac{1}{\sqrt[n]{m}}+\frac{1}{\sqrt[m]{n}}\geq \frac{n + m}{m + n - 1} > 1 \Rightarrow \frac{1}{\sqrt[n]{m}}+\frac{1}{\sqrt[m]{n}} > 1\qquad n, m \in \mathbb{N}$$ How do I prove that $1$ is the greatest lower bound?
Edit (thanks to first comment): Now to prove that it's the greatest bound, by taking any $\epsilon > 0$ and proving that: $$\forall \epsilon > 0 \quad\exists_{m, n \in \mathbb{N}}: \frac{1}{\sqrt[n]{m}}+\frac{1}{\sqrt[m]{n}} > \epsilon + 1$$ Which is, of course, true because for any $\epsilon > 0$ there is a natural number $m$ big enough to make the inequality below true for $n = 1$: $$1 + \frac{1}{m} < 1 + \epsilon$$